66 THE TIDAL PROBLEM. 



we may obtain 



/ pd{ah)==ba,\e,—^)f{a,) / pda' = Z<p{a,) (6), (7) 



from which we obtain 



_47r 



o. = ^Kfia,) (l + -|v[=.-t]) (8) 



.. = ^K.W (l-f [.-f ]) 



(9) 



The relation between m and ^o depends upon the "degree of hetero- 

 geneity" of the earth: that is, it depends on the departure of the surface 

 and central density from the mean density. For a homogeneous earth we 

 may write, as is well known: 



2eo = |m (10) 



but for a body possessing the law of heterogeneity given by equation (2) 

 above, we can deduce the expression * 



The change of the denominator of the fraction on the right-hand side 

 of the equation from 2 to 2.536 is brought about by the change of hy- 

 pothesis from 



( Central density = 5.50 



A J Mean density = 5.50 



t Surface density = 5.50 

 to the hypothesis 



r Central density = 10.74 



B J Mean density = 5.50 



t Surface density == 2.75 



If we further assume that the mean attraction at the surface of the 

 earth is 982 dynes per gram of attracted matter, we may write the equa- 

 tions of equatorial and polar attractions in the simple form 



a« = 982(l-0.1739g (12) 



ap = 982(l + 0.3477£o) (13) 



ilf = 982 (1.0144)^0 (14) 



It is upon the numerical values here written that the results of the 

 following table have been obtained. The results can be checked by sub- 

 stituting for £ in equations (3) and (4) the expression 



£=[0.09645 (a2+ a') + 0.807 l]£o (15) 



which is an algebraic function approximately equivalent to the transcen- 

 dental relation between £ and a, within the interval with which we are 



* See a "Treatise on attractions, Laplace's functions, and the figure of the earth," by- 

 John H. Pratt, London, 1871, p. 116. 



